Q 15. - Calculus

Question

Show that the minimal value of $D(x,y)={\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2$ is $\frac{{\left(a{x}_0+b{y}_0+c{z}_0-d\right)}^2}{a^2+b^2+c^2}$ by evaluating $D\left(\frac{ad-ab{y}_0-ac{z}_0+b^2{x}_0+c^2{x}_0}{a^2+b^2+c^2},\right.\left.\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}\right)$

Step-by-Step Solution

Answer

It can be determined by finding stationary points and discriminant.

Step 1: Given Information

It is given that $D(x,y)={\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2.·$---(1)

Step 2: Differentiate wrt x

Differentiating, we get$\frac{\partial }{\partial x}D(x,y)=\frac{\partial }{\partial x}\left\{{\left(x-x_0\right)}^2+{\left(y-y_0\right)}^2+{\left(\frac{d-ax-by}{c}-z_0\right)}^2\right\}$

$D_x(x,y)=\frac{\partial }{\partial x}{\left(x-x_0\right)}^2+\frac{\partial }{\partial x}{\left(y-y_0\right)}^2+\frac{\partial }{\partial x}{\left(\frac{d-ax-by}{c}-z_0\right)}^2$

$=2\left(x-x_0\right)+0+2\left(\frac{d-ax-by}{c}-z_0\right)\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}-z_0\right)$

$=2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left\{\frac{\partial }{\partial x}\left(\frac{d-ax-by}{c}\right)-\frac{\partial }{\partial x}{z}_0\right\}$

$=2\left(x-x_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(-\frac{a}{c}-0\right)$

$=2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right).$---(2)

Differentiating (1) wrt $y$

$D_y(x,y)=\frac{\partial }{\partial y}{\left(x-x_0\right)}^2+\frac{\partial }{\partial y}{\left(y-y_0\right)}^2+\frac{\partial }{\partial y}{\left(\frac{d-ax-by}{c}-z_0\right)}^2$

$$\begin{gathered} =2\left(y-y_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(\frac{\partial }{\partial y}\frac{d-ax-by}{c}-\frac{\partial }{\partial y}{z}_0\right) \\ =2\left(y-y_0\right)+2\left(\frac{d-ax-by}{c}-z_0\right)\left(-\frac{b}{c}-0\right) \end{gathered}$$

$=2\left(y-y_0\right)-2·\frac{b}{c}\left(\frac{d-ax-by}{c}-z_0\right)$---(3)

Step 3: Stationary Points

It is given by $D_x(x,y)=0\text{ and }{D}_y(x,y)=0$

$D_x(x,y)=0$

$$\begin{gathered} 2\left(x-x_0\right)-2·\frac{a}{c}\left(\frac{d-ax-by}{c}-z_0\right)=0 \\ 2x+\frac{2a^2}{c^2}x-2x_0-\frac{2ad}{c^2}+\frac{2aby}{c^2}+\frac{2a}{c}{z}_0=0 \\ \left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y-2x_0-\frac{2ad}{c^2}+\frac{2a}{c}{z}_0=0 \end{gathered}$$

$\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y=2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\dots$---(4)

Similarly, when $D_y(x,y)=0$

$\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y=2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0$---(5)

Step 4: Calculating y

Multiply (4) by $\frac{2ab}{c^2}$, (5) by $\left(2+\frac{2a^2}{c^2}\right)$ and subtract

$\frac{2ab}{c^2}\left\{\left(2+\frac{2a^2}{c^2}\right)x+\frac{2ab}{c^2}y\right\}-\left(2+\frac{2a^2}{c^2}\right)\left\{\frac{2ab}{c^2}x+\left(2+\frac{2b^2}{c^2}\right)y\right\}$

$=\frac{2ab}{c^2}\left(2x_0+\frac{2ad}{c^2}-\frac{2a}{c}{z}_0\right)-\left(2+\frac{2a^2}{c^2}\right)\left(2y_0+\frac{2bd}{c^2}-\frac{2b}{c}{z}_0\right)$

$$\begin{gathered} =\frac{4ab}{c^2}{x}_0+\frac{4a^{2}bd}{c^4}-\frac{4a^{2}b}{c^3}{z}_0-4y_0-\frac{4bd}{c^2} \\ +\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0-\frac{4a^{2}bd}{c^4}+\frac{4a^{2}b}{c^3}{z}_0 \end{gathered}$$

$\left\{\frac{4a^2{b}^2}{c^4}-\left(2+\frac{2a^2}{c^2}\right)\left(2+\frac{2b^2}{c^2}\right)\right\}y=\frac{4ab}{c^2}{x}_0-4y_0-\frac{4bd}{c^2}+\frac{4b}{c}{z}_0-\frac{4a^2}{c^2}{y}_0$

$$\begin{gathered} \left(\frac{4a^2{b}^2}{c^4}-4-\frac{4b^2}{c^2}-\frac{4a^2}{c^2}-\frac{4a^2{b}^2}{c^4}\right)y \\ =-\frac{4bd}{c^2}+\frac{4ab}{c^2}{x}_0+\frac{4b}{c}{z}_0-4y_0-\frac{4a^2}{c^2}{y}_0 \end{gathered}$$

$-\frac{4}{c^2}\left(a^2+b^2+c^2\right)y=-\frac{4}{c^2}\left(bd-ab{x}_0-bc{z}_0+c^2{y}_0+a^2{y}_0\right)$

$y=\frac{bd-ab{x}_0-bc{z}_0+a^2{y}_0+c^2{y}_0}{a^2+b^2+c^2}$